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Question

Question: \(\sin \left( \cot ^ { - 1 } x \right)\)...

sin⁡(cot⁡−1x)\sin \left( \cot ^ { - 1 } x \right)

A

1+x2\sqrt { 1 + x ^ { 2 } }

B

xx

C

(1+x2)−3/2\left( 1 + x ^ { 2 } \right) ^ { - 3 / 2 }

D

(1+x2)−1/2\left( 1 + x ^ { 2 } \right) ^ { - 1 / 2 }

Answer

(1+x2)−1/2\left( 1 + x ^ { 2 } \right) ^ { - 1 / 2 }

Explanation

Solution

Let cot⁡−1x=θ⇒x=cot⁡θ\cot ^ { - 1 } x = \theta \Rightarrow x = \cot \theta

Now cosec⁡θ=1+cot⁡2θ=1+x2\operatorname { cosec } \theta = \sqrt { 1 + \cot ^ { 2 } \theta } = \sqrt { 1 + x ^ { 2 } }

∴sin⁡θ=1cosec⁡θ=11+x2⇒θ=sin⁡−111+x2\therefore \sin \theta = \frac { 1 } { \operatorname { cosec } \theta } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \Rightarrow \theta = \sin ^ { - 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }

Hence sin⁡(cot⁡−1x)=sin⁡(sin⁡−111+x2)\sin \left( \cot ^ { - 1 } x \right) = \sin \left( \sin ^ { - 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \right)

=11+x2=(1+x2)−1/2= \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } = \left( 1 + x ^ { 2 } \right) ^ { - 1 / 2 }.